JEE Mains · Physics · STD 12 - 10. Wave optics
A ray of light is incident from a denser to a rarer medium. The critical angle for total internal reflection is \({\theta _{iC}}\) and Brewster's angle of incidence is \({\theta _{iB}}\), such that \(\sin \,{\theta _{iC}}/\sin \,{\theta _{iB}} = \eta = 1.28\). The relative refractive index of the two media is
- A \(0.2\)
- B \(0.4\)
- C \(0.8\)
- D \(0.9\)
Answer & Solution
Correct Answer
(C) \(0.8\)
Step-by-step Solution
Detailed explanation
Here, \(\sin \theta_{i c} / \sin \theta_{i B}=1.28\) As we know, \(\mu=\frac{\sin \theta_{\mathrm{iB}}}{\sin \left(\frac{\pi}{2}-\theta_{\mathrm{iB}}\right)}\) where, \(\theta_{\mathrm{iB}}\) is Brewster's angle of incidence, And, \(\mu=\frac{1}{\sin \theta_{\text {ic }}}\)…
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